Maximal domain
The maximal domain is the largest set of real inputs for which every operation in the function is defined. Check each denominator and radicand separately, then intersect the resulting sets
\[\sqrt{R(x)}:R(x)\ge0,\qquad\frac{1}{D(x)}:D(x)\ne0\]
- Exclude denominator zeros A denominator must not equal zero \(\displaystyle \frac{1}{x-4}:x-4\ne0\Rightarrow x\ne4\)
- Include numerator root endpoints A square-root radicand in the numerator may equal zero \(\displaystyle \sqrt{x-2}:x-2\ge0\Rightarrow x\ge2\)
- Make denominator radicands positive A square root in a denominator cannot have a zero radicand \(\displaystyle \frac{1}{\sqrt{x+1}}:x+1>0\Rightarrow x>-1\)
- Intersect all component domains An input must satisfy every restriction in a sum or difference \(\displaystyle [2,\infty)\cap\left(\mathbb{R}\setminus\{5\}\right)=[2,5)\cup(5,\infty)\)
Worked example
Find the maximal real domain of the function below
\[f(x)=\sqrt{3x-6}+\frac{1}{x-5}\]
- \(\displaystyle 3x-6\ge0\Rightarrow3x\ge6\Rightarrow x\ge2\)The numerator square-root radicand must be non-negative
- \(\displaystyle x-5\ne0\Rightarrow x\ne5\)The denominator must be non-zero
- \(\displaystyle x\in[2,\infty)\cap\left(\mathbb{R}\setminus\{5\}\right)=[2,5)\cup(5,\infty)\)Keep only inputs that satisfy both conditions
- \(\displaystyle x=2:\quad3(2)-6=0,\quad2-5=-3\ne0\)The left endpoint is valid because the root radicand is zero and the denominator is non-zero
Common mistakes
A numerator square-root endpoint is made strict
✕\(\displaystyle \sqrt{x+5}\Rightarrow x>-5\)✓\(\displaystyle \sqrt{x+5}\Rightarrow x\ge-5\)
A numerator square-root radicand may equal zeroA denominator zero is allowed
✕\(\displaystyle \frac{1}{x-4}\Rightarrow x\in\mathbb{R}\)✓\(\displaystyle \frac{1}{x-4}\Rightarrow x\ne4\)
Division by zero is undefined, so every denominator zero is excludedDenominator restriction is omitted
✕\(\displaystyle \operatorname{Dom}\left(\sqrt{x-1}+\frac{1}{x-3}\right)=[1,\infty)\)✓\(\displaystyle \operatorname{Dom}\left(\sqrt{x-1}+\frac{1}{x-3}\right)=[1,3)\cup(3,\infty)\)
A sum is defined only where both component conditions hold, so the domains are intersectedDifficulty levels
introFind the domain of one function with a linear denominator or numerator radicand
standardFind restrictions from a factored quadratic denominator or radicand
challengeHandle a root in a denominator or intersect two component restrictions
examSelect the only domain statement satisfying every operation in the function
Questions about this topic
What makes a domain maximal?+
It contains every real input that makes all parts of the function defined, and no additional real input can be included
Why is zero allowed in a numerator square root?+
The square root of zero is defined, so a numerator square root includes the input where its radicand is zero
Why are component domains intersected?+
A sum or difference can be evaluated only when every component can be evaluated at the same input
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 9: Logarithmic calculus
In this chapter
Differentiate natural-logarithmic functions9A Logarithmic differentiation: Evaluate and determine function rules9A Logarithmic differentiation: Find the maximal domain of an algebraic function9A Logarithmic differentiation: Simplify and combine algebraic fractions9A Logarithmic differentiation: Expand and factor polynomial expressions9A Logarithmic differentiation: Evaluate a polynomial or recover a coefficient parameterApply logarithmic derivatives9B Applications of logarithmic derivatives: Form and analyse an inverse relation or functionFind the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular line9B Applications of logarithmic derivatives: Solve a linear equation9B Applications of logarithmic derivatives: Evaluate a polynomial or recover a coefficient parameter9B Applications of logarithmic derivatives: Solve a higher-degree polynomial equation or inequality9B Applications of logarithmic derivatives: Find the maximal domain of an algebraic function9B Applications of logarithmic derivatives: Evaluate and determine function rulesAnti-differentiate reciprocal and linear-reciprocal functions9C Reciprocal and linear-reciprocal antiderivatives: Simplify and combine algebraic fractions9C Reciprocal and linear-reciprocal antiderivatives: Expand and factor polynomial expressions9C Reciprocal and linear-reciprocal antiderivatives: Solve a linear equationIntegration by recognition with logarithmic components9D Recognising logarithmic derivatives: Expand and factor polynomial expressions9D Recognising logarithmic derivatives: Round to decimal places and significant figuresFind areas involving reciprocal curvesFind the area under a logarithmic curve using its inverseLogarithmic models in context9E Areas, inverse functions and logarithmic models: Form and analyse an inverse relation or function9E Areas, inverse functions and logarithmic models: Solve a linear equation9E Areas, inverse functions and logarithmic models: Solve a higher-degree polynomial equation or inequalityPerimeter and area of standard shapes9E Areas, inverse functions and logarithmic models: Round to decimal places and significant figures